Two electric bulbs, rated at (25 W, 220 V) and (100 W, 220 V), are connected in series across a 220 V voltage source. If the 25 W and 100 W bulbs draw powers $$P_1$$ and $$P_2$$ respectively, then:

For every electric bulb the stated rating, for example “(25 W, 220 V)”, tells us that when a potential difference of 220 V is applied across its terminals it consumes 25 W of power. Using the basic electrical relation $$P=\dfrac{V^{2}}{R}\,,$$ we can obtain the resistance of each filament at its working temperature.

For the 25 W bulb we have

$$R_{25}=\dfrac{V^{2}}{P}=\dfrac{(220)^{2}}{25} =\dfrac{48400}{25}=1936\;\Omega.$$

For the 100 W bulb we write

$$R_{100}=\dfrac{V^{2}}{P}=\dfrac{(220)^{2}}{100} =\dfrac{48400}{100}=484\;\Omega.$$

The two bulbs are now connected in series across the same 220 V supply, so their resistances simply add. The total series resistance is therefore

$$R_{\text{series}}=R_{25}+R_{100}=1936+484=2420\;\Omega.$$

The same current flows through both resistances in series. Ohm’s law gives this current as

$$I=\dfrac{V_{\text{source}}}{R_{\text{series}}} =\dfrac{220}{2420}\;{\rm A} =\dfrac{1}{11}\;{\rm A}\approx 0.090909\;{\rm A}.$$

Now, to find the actual power dissipated in each bulb under this new condition, we employ the formula $$P=I^{2}R.$$ For the 25 W bulb we calculate

$$P_{1}=I^{2}R_{25} =\left(\dfrac{1}{11}\right)^{2}\times 1936 =\dfrac{1}{121}\times 1936 =\dfrac{1936}{121} =16\;{\rm W}.$$

Next, for the 100 W bulb we write

$$P_{2}=I^{2}R_{100} =\left(\dfrac{1}{11}\right)^{2}\times 484 =\dfrac{1}{121}\times 484 =\dfrac{484}{121} =4\;{\rm W}.$$

Thus under series connection the 25 W bulb actually consumes $$P_{1}=16\;{\rm W}$$ while the 100 W bulb consumes only $$P_{2}=4\;{\rm W}.$$

Comparing with the given choices, we see that this corresponds to Option D.

Hence, the correct answer is Option D.